The Interior of a Set of Complex Numbers

# The Interior of a Set of Complex Numbers

 Definition: Let $A$ be a set of complex numbers. The Interior of $A$, denoted $\mathrm{int} (A)$, is defined to be the largest open set contained in $A$.

By "largest open set contained in $A$" we mean that any other open set contained in $A$ is also contained in $\mathrm{int} (A)$.

We will now prove some properties about the interior of a set of complex numbers.

 Theorem 1: Let $A$ be a set of complex numbers. Then $A$ is open if and only if $A = \mathrm{int} (A)$.
• Proof: $\Rightarrow$ Suppose that $A$ is open. Then clearly the largest open set contained in $A$ is $A$ itself, so $A = \mathrm{int} (A)$.
• $\Leftarrow$ Suppose that $A = \mathrm{int}(A)$. By definition, $\mathrm{int}(A)$ is the largest open set contained in $A$, and so $\mathrm{int}(A) = A$ is open. $\blacksquare$
 Theorem 2: Let $A$ and $B$ be sets of complex numbers. Then: a) $\mathrm{int}(A) \cup \mathrm{int}(B) \subseteq \mathrm{int}(A \cup B)$. b) $\mathrm{int}(A) \cap \mathrm{int}(B) = \mathrm{int}(A \cap B)$.

Observe that $\mathrm{int}(A \cup B)$ is not contained in $\mathrm{int}(A) \cup \mathrm{int}(B)$ in general. For example, let $A = (0, 1]$ and let $B = (1, 2]$. Then $\mathrm{int}(A) = (0, 1)$ and $\mathrm{int}(B) = (1, 2)$. So $\mathrm{int}(A) \cup \mathrm{int}(B) = (0, 1) \cup (1, 2)$.

Now $A \cup B = (0, 2]$ and $\mathrm{int}(A \cup B) = (0, 2)$.

Clearly $\mathrm{int}(A) \cup \mathrm{int}(B) \not \supseteq \mathrm{int}(A \cup B)$.